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Question: Evaluate $\int \frac{2z+3}{z} dz$ over the curve C
0

Where C is,

(i) upper half of the circle |z|=2

(ii) lower half of the circle |z|=2

(iii) whole circle in anticlockwise direction

Subject: Applied Mathematics 4

Topic: Complex Integration

Difficulty: Medium

m4e(34) • 89 views
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modified 18 days ago  • written 3 months ago by gravatar for Manan Bothra Manan Bothra0
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(i) Upper half of |z| = 2

|z| = 2; z = 2e$^{i \theta}$

dz = 2i$^{i \theta} \,\, d \theta $

enter image description here

$ \int_c \frac{2z + 3}{2} \,\, dz \\ = \int_0^{\pi} \frac{2(2^{i \theta})}{2^{i \theta}} 2i^{i \theta} \,\, d \theta \\ = i \int_0^{\pi} (4e^{i \theta} + 3) \,\, d\theta \\ = i [\frac{4 e^{i \theta}}{i} + 3 \theta]_0^{\pi} \\ = i [\frac{4e^{i \pi}}{i} - \frac{4e^0}{i} + 3\pi] \\ = i [\frac{-4}{i} - \frac{4}{i} + 3 \pi] \\ = i[\frac{-8}{i} + 3 \pi] \\ = 3 \pi i - 8 $

(ii) Lower half of |z| = 2

enter image description here

$ |z| = 2; \\ z = 2e^{i \theta}; \\ dz = 2ie^{i \theta} \,\, d \theta \\ \int_{\pi}^{2\pi} \frac{2(2e^{i \theta} + 3)}{2e^{i \theta}}(2ie^{i \theta}) \,\, d \theta \\ i \int_{\pi}^{2\pi}(4e^{i \theta} + 3) \,\, d\theta \\ = i[\frac{4e^{i \theta}}{i} +3 \theta]_{\pi}^{2\pi} \\ = i (\frac{8+3\pi i}{i}) \\ = 3 \pi i + 8 $

(iii) Whole circle in anti-clockwise direction

enter image description here

$ \theta = 0 \to 2\pi $

$ z = 2e^{i \theta} \\ z = 2ie^{i \theta} \,\, d\theta \\ \int_{0}^{2\pi} \frac{2(2e^{i \theta} + 3)}{2e^{i \theta}}(2ie^{i \theta}) \,\, d \theta \\ = i \int_{0}^{2\pi}(4e^{i \theta} + 3) \,\, d\theta \\ = i[\frac{4e^{i \theta}}{i} +3 \theta]_{0}^{2\pi} \\ = i [\frac{4(1)}{i} + 6 \pi - \frac{4}{i}] \\ = 6 \pi i $

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written 18 days ago by gravatar for Manan Bothra Manan Bothra0
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